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11 Two and many electron atoms 149 11 Two and many electron atoms 11.1 Helium z m, -e r12 r1 m, -e θ1 θ2 r2 M, +2e φ2 y φ1 x Abbildung 11.1. Helium atom with two electrons 1 and 2 at positions r2 and r2 . The partial dierential equation of a three particle system is not solvable. In Helium we have one nucleus and two electrons and the SE is given by · ¸ ~2 ~2 2e2 2e2 e2 → → → → − 41 − 42 − − + ψE (− r 1, − r 2 ) = EψE (− r 1, − r 2 ). 2m 2m 4πε0 r1 4πε0 r2 4πε0 r12 (11.1) Because of the presence of the electron-electron interaction term 1/r12 this equa→ → tion is not separable, so that an eigenfunction ψ(− r 1, − r 2 ) of (11.1) cannot be written in the form of a single product of one-electron wave functions. The wave functions are said to be entangled. We remark that the equation (11.1) is un→ − changed when the coordinates of the two electrons are interchanged (− r1↔→ r 2 ). Thus, if we denote by P12 an interchange operator that permutes the spatial coordinates of the two electrons, the wave functions → → → → P12 ψ(− r 1, − r 2 ) = ψ(− r 2, − r 1) (11.2) satisfy the same SE. If both functions solve the same SE with non degenerate Physics of Atoms and Molecules 150 11 Two and many electron atoms eigenvalues E the functions can only dier by a factor λ. → → → → → − P12 ψ(− r 1, − r 2 ) = ψ(− r 2, − r 1 ) = λψ(− r 1, → r 2) − → − → − → → − − → − 2 2 P12 ψ( r 1 , r 2 ) = λ ψ( r 1 , r 2 ) = ψ( r 1 , → r 2) − → − → − → → − ψ( r 2 , r 1 ) = ±ψ( r 1 , r 2 ) (11.3) (11.4) (11.5) → → − → Moreover, both functions ψ(− r 1, − r 2 ) and → r 2 , ψ(− r 1 ) must be continuous, singlevalued and bounded. The eigenvalues of P12 are ±1. The eigenfunctions are space− → → → symmetric ψ+ (→ r 1, − r 2 ) (λ = 1) or space-antisymmetric ψ− (− r 1, − r 2 ) (λ = −1). States described by space-symmetric wave functions are called para states; those corresponding to space-antisymmetric wave functions are known as ortho states. For the parity operator we have [H0 , P12 ] = 0 (11.6) and the quantum number of the parity operator (+1, −1) are good quantum numbers describing the space-symmetry of the wave function. Experimentally we can acquire information on Helium atoms by ionizing the Helium atom. 24.6 eV : Ionization of the rst electron 54.4 eV : Ionization of the second electron 79.0 eV : Sum of both, total energy of both electrons The energy of 54.4 eV is known, since we already solved the SE for atoms like He+ to be E1 = − Z 2 ~2 ~2 = −4 = −4 × 13.6 eV = −54.4 eV 12 2ma20 2ma20 (11.7) The smaller binding energy of the rst released electron is due to the shielding ef2 fect of the remaining inner electron. If we take 4πεe0 r12 as a perturbation H 0 we can apply perturbation theory. The energy without the electron-electron interaction term is H0 = − ~2 ~2 2e2 2e2 41 − 42 − − = −108.8 eV. 2m 2m 4πε0 r1 4πε0 r2 (11.8) The dierence between -79 eV and 108.8 eV is not small so that we can expect the result of the perturbation theory not to be very precise. The unperturbed Hamiltonian is 2 X ~2 2e2 − 4i − . (11.9) H0 = H1 + H2 = 2m 4πε0 ri i=1 Physics of Atoms and Molecules 11.1 Helium 151 Equation (11.9) is the sum of two one-electron hydrogenic Hamiltonians. For such → → a decoupled Hamiltonian we can write the wave function ψ(− r 1, − r 2 ) in the form of single products of hydrogenic wave funtions → − − → ψ (0) (− r 1, → r 2 ) = ψa ( → r 1 )ψb (− r 2 ). (11.10) Where we abbreviated a = n1 `1 m1 and b = n2 `2 m2 . The corresponding discrete energies being given by → − H0 ψ (0) (− r 1, → r 2) = = = = = → − H0 ψa (− r 1 )ψb (→ r 2) (11.11) − → − → (H1 + H2 )ψa ( r 1 )ψb ( r 2 ) (11.12) − → → − − → − → ψb ( r 2 )H1 ψa ( r 1 ) + ψa ( r 1 )H2 ψb ( r 2 ) (11.13) − → − → − → → − ψb ( r 2 )Ea ψa ( r 1 ) + ψa ( r 1 )Eb ψb ( r 2 ) (11.14) − → → − → → − (0) − (Ea + Eb )ψa ( r 1 )ψb ( r 2 ) = Eψ ( r 1 , r 2 ). (11.15) In general for N 'uncoupled' electrons we have N single electron Hamiltonians H0 = N X (11.16) Hi i → → Hi ψai (− r i ) = Eai ψai (− r i) N Y → ψ = ψai (− r i) (11.17) (11.18) i E = X i E0 = −4 (11.19) Eai µ ~2 2ma20 ¶· ¸ 1 1 + + ... . n21 n22 (11.20) Using this approximation we nd E0 to be E0 = −108.0eV. We still have to take into account the symmetry of the wave function and the electron-electron interaction term. For the electron-electron interaction term we nd in rst order perturbation theory Z Z e2 → → → → − → r 2) ψa (− r 1 )ψb (− r 2 )d− r 1 d→ r 2 (11.21) r 1 )ψb∗ (− ∆E = ψa∗ (− 4πε0 r12 Z Z e2 1 → → − → = |ψa (− r 1 )|2 − |ψb (− r 2 )|2 d→ r 1 d− r2 (11.22) → → 4πε0 |r 2−− r 1| ≡ J (11.23) The integral J is called the Coulomb integral describing the electrostatic inter→ action energy of two overlapping distributions of charge density %n1 (− r 1 ) and → %n2 (− r 2 ). With the denition of charge densities we can rewrite the Coulomb Physics of Atoms and Molecules 152 11 Two and many electron atoms integral → → r i ) = −e|ψni (− r i )|2 %ni (− Z Z → → 1 %n1 (− r 1 )%n2 (− r 2) − → Jn1 n2 = d→ r 1 d− r 2. − → − → 4πε0 | r 2 − r 1| (11.24) (11.25) Now, we evaluate J1s1s for the unperturbed wave functions to get the energy correction ∆E . With µ 3 ¶1/2 Z → − ψ1s ( r i ) = e−Zri /a0 (11.26) πa30 ∞ X̀ ` X 1 4π r< = Y ∗ (ϑ , ϕ1 )Y`m (ϑ2 , ϕ2 ) (11.27) `+1 `m 1 r12 2` + 1 r > `=0 m=−` where in equation (11.27) r< is the smaller r of the two radial distances r1 and r2 , whereas r> is the bigger r of the two radial distances r1 and r2 , we have e2 J = 4πε0 µ Z3 πa30 ¶2 Z Z −2 e Z(r1 +r2 ) a0 ∞ X̀ X ` 4π r< × `+1 2` + 1 r > `=0 m=−` ∗ Y`m (ϑ1 , ϕ1 )Y`m (ϑ2 , ϕ2 )r12 r22 dr1 dr2 d cos ϑ1 d cos ϑ2 dϕ1 dϕ(11.28) 2 All values for ` 6= 0 and m 6= 0 vanish, because of the orthornormality condition of the spherical harmonics and we have e2 J = (4π)2 4πε0 2 = e (4π)2 4πε0 µ µ Z3 πa30 3 Z πa30 ¶2 Z∞ Z∞ −2 e ¶2 0 0 Z∞ r22 e 0 | Zr1 a0 e −2  Z(r ) −2 a 2 0 Zr2 a0 Zr2  0 r12 r22 dr1 dr2 r> −2 r12 e Z(r1 ) a0 r2 (11.29) Z∞ dr1 + {z r2  1) r12 −2 Z(r a0 r1 e dr1  dr2 } 5a5 0 27 Z 5 (11.30) 2 = 5 e Z = 34.0 eV. 8 4πε0 a0 (11.31) Thus the energy is EHe = 2E1s + J = −108.8 + 34.0 = −74.8 eV. (11.32) This is close to the experimentally measured energy of −79.0 eV and is only 5% o. Physics of Atoms and Molecules 11.2 Excited states of helium 153 11.2 Excited states of helium If one of the two electrons in helium is not in its ground state but in a higher orbital, we nd the energy of the excited state in zero order perturbation to be µ ¶ 1 (0) 2 E1,n = −Z 13.6 eV 1 + 2 . (11.33) n The energy levels corresponding to the wave functions are degenerate in ` and m and also exhibit the exchange degeneracy. Upon exchanging the electrons 1 → → and 2 we nd the same energy for the two wave functions ψa (− r 1 )ψb (− r 2 ) and − → ψa (→ r 2 )ψb (− r 1 ). We can solve this problem in two ways: With perturbation theory on degenerate levels we have to solve ¯ ¯ ¯ H11 − ES11 H12 − ES12 ¯ ¯ ¯ ¯ H21 − ES21 H22 − ES22 ¯ = 0. (11.34) with Hij Sij i j = = = = hi|H|ji hi|ji → → ψ1 = ψa (− r 1 )ψb (− r 2) − → − → ψ2 = ψa ( r 2 )ψb ( r 1 ). (11.35) (11.36) (11.37) (11.38) Evaluating of equation (11.34) leads to H11 S11 S12 H12 κab H22 = Ea + Eb + Jab S22 = 1 S21 = 0 H21 = κab Z Z → → = ψa∗ (− r 1 )ψb∗ (− r 2) (11.39) (11.40) (11.41) (11.42) = = = = e2 → → → − ψa (− r 2 )ψb (− r 1 )d− r 1 d→ r 2 . (11.43) 4πε0 r12 The integral κab in equation (11.43) is called exchange integral and is a pure quantum mechanical eect of two particles that cannot be distinguished. The exchange integral can also be represented by the exchange charge density − → → % (→ r ) = −eψ ∗ (− r )ψ (− r ). (11.44) ab i a 1 b 1 → The exchange integral is a measure of the overlap of the wave functions ψa (− r 1) → and ψb (− r 1 ). If the overlap vanishes there is no exchange degeneracy and the integral is zero (κab = 0). With the relations above we can solve equation (11.34) ¯ ¯ ¯ ¯ Ea + Eb + Jab − E κab ¯ = 0. ¯ (11.45) ¯ κab Ea + Eb + Jab − E ¯ E± = Ea + Eb + Jab ± κab . (11.46) Physics of Atoms and Molecules 154 11 Two and many electron atoms So the energy levels are split by the exchange energy κab with the normalized wave functions given by 1 − → → → ψ± = √ [ψa (→ r 1 )ψb (− r 2 ) ± ψa (− r 2 )ψb (− r 1 )] . 2 (11.47) An alternative approach is conclude from the application of the parity operator that the wave functions have to be space-symmetric or space-antisymmetric. − → → → ψs ( → r 1, − r 2 ) = ψ+ (− r 1, − r 2) 1 − → → → = √ [ψa (→ r 1 )ψb (− r 2 ) + ψa (− r 2 )ψb (− r 1 )] 2 − → → → ψas (→ r 1, − r 2 ) = ψ− (− r 1, − r 2) 1 − → → − = √ [ψa (→ r 1 )ψb (− r 2 ) − ψa (− r 2 )ψb (→ r 1 )] 2 (P ara) (11.48) (Ortho).(11.49) Without any calculations we conclude that e2 h+|H |−i = h+| |−i = 0 = 4πε0 r12 0 Z → → ψs∗ H 0 ψas d− r 1 d− r 2. (11.50) In the basis of ψs and ψas the perturbation H 0 is already diagonal. ∆E+ = hψs |H 0 |ψs i (11.51) 0 ∆E− = hψas |H |ψas i (11.52) Z 2 1 e ∗ − → → − ∆E± = {ψa (→ r 1 )ψb (− r 2 ) ± ψa (− r 2 )ψb (→ r 1 )} × 2 4πε0 r12 → → − → → − {ψa (− r 1 )ψb (− r 2 ) ± ψa (→ r 2 )ψb (− r 1 )} d− r 1 d→ r2 (11.53) Z 2 1 e → → → → = |ψa (− r 1 )|2 |ψb (− r 2 )|2 d− r 1 d− r2 2 4πε0 r12 Z 1 e2 → → → → + |ψa (− r 2 )|2 |ψb (− r 1 )|2 d− r 1 d− r2 2 4πε0 r12 Z 1 e2 → → → → → → ± ψa∗ (− r 1 )ψa (− r 2) ψ ∗ (− r 2 )ψb (− r 1 ) d− r 1 d− r2 2 4πε0 r12 b Z 1 e2 → → → → → → ± r 2 )ψa (− r 1) r 1 )ψb (− r 2 ) d− r 1 d− r2 ψa∗ (− ψ ∗ (− 2 4πε0 r12 b (11.54) Physics of Atoms and Molecules 11.3 Indistinguishable particles 155 The rst two and the last two integrals in equation (11.54) are equal so that ∆E± becomes Z e2 − − → − ∆E± = |ψa (→ r 1 )|2 |ψb (→ r 2 )|2 d− r 1 d→ r2 4πε0 r12 | {z } Z ± Jab → → ψa∗ (− r 2 )ψa (− r 1) | e2 → → − → ψb∗ (− r 1 )ψb (− r 2 ) d→ r 1 d− r2 4πε0 r12 {z } (11.55) κab (11.56) ∆E± = Jab ± κab The twofold degenerated levels are shifted (Jab ) and split by the energy spacing of 2κab , which is a result of the electron-electron interaction. The energy levels of a) b) +Κab −Κab Ea + Eb | ψ− | 2 ψ+ ψ− Jab r1 - r2 Abbildung 11.2. a) Shifting and splitting of the energy levels due to electrostatic interaction (Jab ) and exchange interaction (κab ). b) Probability of nding two electrons of the antisymmetric wave function at the same position. The probability is zero for r12 = 0; this is called Fermi-hole. the antisymmetric wave functions ψ− exhibit higher negative energies (Ea + Eb + Jab − κab ), because the electrons omit the same space and therefore the repulsion is smaller. The values of Jab and κab are always positive. In the ground state with a = b the antisymmetric wave function is constant zero, so that the wave function of the ground state is space-symmetric ψ+ with energy E = 2Ea + Jaa . Excited states can be either space-symmetric or space-antisymmetric and there are no allowed optical transitions in the dipole approximation between para (ψ+ ) and ortho (ψ− ) states. Z − → → → → → − → → → r +− r ] ψ (− r ,− r ) d− r 1 d− r2=0 (11.57) h µ s,as i = −e ψs (− r 1, − r ) [→ {z 2} | 1 {z 2} | as {z1 2} | + + − Optical transitions between ψ− and ψ+ are forbidden (exception: strong LS coupling, intercombination lines). Ortho helium cannot be transferred to para helium by light absorption (behave like dierent atoms). 11.3 Indistinguishable particles Two particles are said to be identical when they cannot be distinguished by means of any intrinsic property. While in classical physics the particles can be Physics of Atoms and Molecules 156 11 Two and many electron atoms distinguished by their paths of sharp trajectories, this is not possible in quantum mechanics, if the particles are found in the same regions of space such as their interaction region. Let us consider a quantum mechanical system of N identical particles. The Hamiltonian and all observables corresponding to this system must be symmetric with respect to any interchange of the space and spin coordinates of the particles (i and j ), dened by the parity operator Pij . (11.58) [Pij , H] = 0. Note, if there are more than N = 2 particles we have N ! permutations and P do not commute among themselves (Pij , Pjk ). There are two exceptional states which are eigenstates of H and of the N ! permutation operators P , the totally symmetric state ψs (q1 , . . . , 1N ) and the totally antisymmetric state ψas (q1 , . . . , 1N ) P ψs (q1 , . . . , 1N ) = ψs (q1 , . . . , 1N ) P ψas (q1 , . . . , 1N ) = ψas (q1 , . . . , 1N ) × ½ (11.59) 1, even permut. −1, odd permut. (11.60) Note, equation (11.58) implies that P is a constant of the motion so that the system of identical particles represented by ψs or ψas will keep that symmetry at all times. To our present knowledge of particles occurring in nature, the two types of states ψs and ψas are thought to be sucient to describe all systems of identical bosons and identical fermions, respectively. This is called the symmetrization postulate. Bosons (mesons, photons) are particles of zero or integer spin, described by totally symmetric wave functions and obey Bose-Einstein statistics. Fermions are particles having half-odd integer spin (electrons, protons, neutrino) and are described by totally antisymmetric wave functions. They satisfy Fermi-Dirac statistics. 11.4 Wave functions of helium The correct wave function describing the electrons in the helium atom has to be a totally antisymmetric wave function, since the electrons are fermions. In the case of negligible spin-orbit coupling the Hamiltonian has no terms coupling space coordinates and spin coordinates resulting in a product type wave function → → → → ψ(− r 1 , ms1 ; − r 2 , ms2 ) = ψ(− r 1, − r 2 )χ(ms1 , ms2 ) (11.61) which has to be antisymmetric by interchanging all coordinates of two electrons. There are two possibilites to full this condition → → ψ− (− r 1 , ms1 ; − r 2 , ms2 ) = ½ → → ψ+ (− r 1, − r 2 )χ− (ms1 , ms2 ) − → − → ψ− ( r 1 , r 2 )χ+ (ms1 , ms2 ) Physics of Atoms and Molecules (11.62) 11.4 Wave functions of helium 157 Also the spin wave function χ(ms1 , ms2 ) can be written in the product form, (in the absence of spin-spin coupling) with four dierent combinations χ(ms1 , ms2 ) χ1 (1, 2) χ2 (1, 2) χ3 (1, 2) χ4 (1, 2) = = = = = (11.63) (11.64) (11.65) (11.66) (11.67) χ(ms1 )χ(ms2 ) α(1)α(2) ↑↑ α(1)β(2) ↑↓ β(1)α(2) ↓↑ β(1)β(2) ↓↓ These four combinations are not symmetrized but upon introducing the total Spin − → − → → − operator S = S 1 + S 2 we can achieve a symmetrized new basis by linear combi− → − → − → → − nation of these four wave functions, that are eigenfunctions of S 2 , S z , S 21 , S 22 and P12 . χ1,1 (1, 2) = α(1)α(2) (11.68) ↑↑ 1 χ1,0 (1, 2) = √ [α(1)β(2) + β(1)α(2)] 2 χ1,−1 (1, 2) = β(1)β(2) ↓↓ 1 χ0,0 (1, 2) = √ [α(1)β(2) − β(1)α(2)] 2 ↑↓ + ↓↑ (11.69) (11.70) ↑↓ − ↓↑ (11.71) − → → − The indices in χS,ms are the quantum numbers of S 2 and S z . The three functions {χ1,1 , χ1,0 , χ1,−1 } have a total spin of S = 1 and magnetic spin quantum numbers of ms = 1, 0, −1. These three spin wave functions represent the symmetric spin wave functions, with total spin S = 1 and belong to Spin Triplet states. The other spin wave function χ0,0 is the antisymmetric spin wave function with total spin S = 0 and belong to Spin Singlet states. Hence the correct wave functions are 1 → → ψ+ (− r 1, − r 2 ) = √ [α(1)β(2) − β(1)α(2)] (11.72) 2 − → for para or singlet states (| S | = 0, spins antiparallel or zero) or    α(1)α(2)  − → √1 [α(1)β(2) + β(1)α(2)] ψ− ( → r 1, − r 2) ×  2  β(1)β(2) (11.73) − → for ortho or triplet states (| S | 6= 0). The Pauli principle says that there are no two identical particles (fermions) with the same quantum numbers. Therefore, we have a correlation between the electrons resulting in specic allowed combinations of spin and space wave functions. The ground state of helium has a symmetric space wave function and consequently an antisymmetric spin function so that the Physics of Atoms and Molecules 158 11 Two and many electron atoms ground state of helium is a singlet state. The excited states of helium (10)(n`) are split by ± E10,n` = E10 + En` + J10,n` ± κ10,n` Z∞ Z∞ r2 2 2 J10,n` = Rn` (r2 )r22 R10 (r1 ) 1 dr1 dr2 r> 0 κ10,n` 1 = 2` + 1 (11.74) (11.75) 0 Z∞ Z∞ R10 (r2 )Rn` (r2 )r22 0 0 ` r12 r< R10 (r1 )Rn` (r1 ) `+1 dr1 dr2 (11.76) r> with para-He (+) and ortho-He (-). The triplet states are energetically lower than the singlet states, due to the reduced electron-electron repulsion. The triplet state (1s)(2s) is metastable, since relaxation to (1s)2 is not possible upon light emission (intercombination lines are forbidden). Excitation of triplet states is possible upon collisions of atoms, spin-orbit coupling (which is very weak for helium) and magnetic dipole transitions. Interaction with an electromagnetic eld in the dipole approximation do not result in changing the total electron spin, because for such interactions it is ∆S = 0 (11.77) and the transition is spin forbidden. Both integrals (11.75, 11.76) depend explicitly on the quantum number `. Thus, degeneracy in m` is still present but not in `. For a given n the Coulomb interaction J10,n` increases with increasing `. This can be explained with bigger Coulomb repulsion upon decreasing hrn` i with increasing `. − → − → − → S = S1+ S2 (11.78) → −2 − →2 − →2 → − − → S = S 1 + S 2 + 2S 1 · S 2 (11.79) · ¸ 2 → − − → 3 ~ s(s + 1) − χsms (11.80) ( S 1 · S 2 )χsms = 2 2 ( − → − → ) 1 S · 1 S2 ± E10,n` = E10 + En` + J10,n` − 1+4 κ10,n` 2 ~2 (11.81) The energy splitting in equation (11.81) can be interpreted as an additional force called exchange force. The sign of the exchange force depend on the orientation of the spins, and the strengths of the force is sucient to orient neighboring spins (ferromagnetism). The naming of the dierent states are as follows n2S+1 LJ (11.82) n is the principal quantum number; S is the total spin quantum number (for helium 0 and 1); L is the total orbital angular momentum quantum number (S, P, D, F, ...); J is the total angular momentum quantum number. Physics of Atoms and Molecules 11.5 Slater determinant 159 2 1P 1s2p Κ2,1 J2,1 2 3P 2 1S 1s2s Κ2,0 J2,0 2 3S (0) E1,2 Abbildung 11.3. The splitting of the unperturbed helium level for n = 2 by the Coulomb integrals and the exchange integrals. 11.5 Slater determinant The total wave function is antisymmetric upon exchange of all coordinates of two electrons. If there are N independent electrons the wave function can be written as a product of N single electron wave functions φi . To full the symmetry conditions of the wave function, the wave function can be written as a sum of Slater determinants ¯ ¯ ¯ φ1 (1) φ1 (2) . . . φ1 (N ) ¯ ¯ ¯ ¯ ¯ φ (1) φ (2) . . . φ (N ) 1 ¯ 2 2 2 ¯ ψas (1, 2, . . . , N ) = √ ¯ (11.83) ¯. .. .. .. .. ¯ . . . . N ! ¯¯ ¯ ¯ φN (1) φN (2) . . . φN (N ) ¯ Using the formalism of the slater determinant, we can write the ground state wave function of helium: φ1 (1) φ1 (2) φ2 (1) φ2 (2) = = = = 1s(1)α(1) 1s(2)α(2) 1s(1)β(1) 1s(2)β(2) ¯ ¯ 1 ¯¯ 1s(1)α(1) 1s(2)α(2) ¯¯ ψas (1, 2, . . . , N ) = √ ¯ 2 1s(1)β(1) 1s(2)β(2) ¯   1 = √ 1s(1)1s(2) [α(1)β(2) − α(2)β(1)] {z } 2 | {z } | + (11.84) (11.85) (11.86) (11.87) (11.88) (11.89) − The Slater determinant and sums of Slater determinants give always antisymmetric wave functions Physics of Atoms and Molecules 160 11 Two and many electron atoms Upon exchange of two rows or two columns, which correspond to exchange of all coordinates of two electrons Pij , the Slater determinant change its sign. If the quantum numbers of two electrons are the same, two rows are equivalent and the Slater determinant is zero. That correspond to the Pauli exclusion principle that no two electrons in a system can have the same quantum numbers. 11.6 Many electron atoms The SE for atoms with N electrons is " N N X ~2 X Ze2 → → → Eψ(− r 1, − r 2, . . . , − r N) = − 4i − 2m 4πε0 ri i i  N N 1 e2 X X  − → → − − → +  ψ( r 1 , r 2 , . . . , r N ). − → → 4πε |r −− r | 0 i j i<j i j (11.90) With increasing Z the electron-electron repulsion becomes stronger and stronger and cannot be treated as a perturbation with respect to the electron-nucleus interaction. For N electrons we have 1/2N (N − 1) electron-electron interaction terms. Hartee (1928) used the averaged electron-electron interaction, which is for the j th electron → → %j (− r j ) = −e|φj (− r j )|2 . (11.91) and postulated a SE for single electron wave functions to be " N # N 2 X ~2 X X Ze − → → ²i φi (→ r i) = − 4i − −e Vj (− r i ) φi (− r i ) (11.92) 2m 4πε r 0 i i i j6=i Z − → 2 |φj ( r j )| − e − Vj (→ r i) = − d→ r j. (11.93) − − 4πε |→ r −→ r | 0 i j The Hartree-Fock equations add an exchange term to the Hartree equation. # " N N → − 2 2 XZ 2 X X ~2 Ze e |φ ( r )| j j − → → 4i − + d→ r j φi (− r i) ²i φi (− r i) = − → − → − 2m 4πε r 4πε | r − r | 0 i 0 i j i i j6=i ½Z ∗ − ¾ → r j )φi (− r j) − φj ( → e2 X → → − d r j φj (− r i ). (11.94) → − → − 4πε0 | r i − r j| j6=i Solutions to the Hartree equations are found iteratively. One starts with zero order single electron wave functions and potentials, calculates the energies, sort Physics of Atoms and Molecules 11.6 Many electron atoms 161 the energies obeying the Pauli principle and determine the averaged potential for the ith electron. With the help of the new potential one calculates the new wave function. If this wave function is the same wave function as before the calculation is nished. If the wave function has changed, it is used for sorting the energies and so on. The solution is called a self-consistent solution. From the calculated energies one can determine the congurations with the lowest energies and ll up the orbitals with electrons. This gives the building up of atoms. Since there is no ` degeneracy due to the Coulomb and exchange integral and the energy increases with increasing ` the orbitals are lled up in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p (11.95) The underlined states have similar energies and lead to complex transition spectra. Note, E4s < E3d due to the higher probability density of 4s electrons at r = 0 than for 3d electrons, leading to an eectively increased nucleic charge for 4s electrons compared to 3d electrons (but hri3d < hri4s ). The wave functions of many electron atoms are similar to single-electron wave functions with the same Y (ϑ, ϕ), but with changed Rn` . The quantum numbers are the same n, `, m` , ms . Electrons with the same quantum numbers n and ` belong to the same subshell (n`). Those electrons are called equivalent electrons. Due to the Pauli exclusion principle there are maximally 2(2` + 1) electrons per (n`) subshell. s ` = 0 2` + 1 = 1 δi = 2(2` + 1) = 2 p 1 3 6 d 2 5 10 f 3 7 14 g 4 9 18 (11.96) (11.97) (11.98) (11.99) The degeneracy di is given by di = δi ! νi !(δi − νi )! (11.100) the possible positions and the number of identical electrons. For example two electrons (νi ) in a (2p) subshell have degeneracy 15. di = 6! = 15. 2!4! (11.101) Now, we can build up the many electron atoms by lling up the subshells according to their energies and obeying the Pauli exclusion principle. Electrons in an open shell have the highest energy and are most weakly bound. They are called − → valence electrons. Closed shells exhibit a total orbital angular momentum L and − → total spin angular momentum S of zero. In the many electron atom nomenclature closed shells are described by 1 S0 (L = 0, S = 0, J = 0). They have a radial Physics of Atoms and Molecules 162 11 Two and many electron atoms Abbildung 11.4. Ionization potential for neutral atoms as a function of nucleic charge Z. symmetric charge distribution and are chemically inert (nobel gases: He, Ne, Ar, Kr, Xe). Alkali atoms with one valence electron in the s-orbital have low binding energy, low ionization energy, increased volume and are chemical reactive. We see that the degeneracy is high in m` and ms . In the case of non equivalent electrons in dierent subshells the Pauli exclusion principle is fullled and we nd for two electrons ` = |`1 − `2 |, |`1 − `2 | + 1, . . . , `1 + `2 s = |s1 − s2 |, |s1 − s2 | + 1, . . . , s1 + s2 . (11.102) (11.103) For example (np)(n0 p) gives 6 dierent terms `1 = `2 = 1 1 s1 = s2 = 2 L = 0, 1, 2 S = 0, 1 ⇒ 1 S, 1 P, 1 D, 3 S, 3 P, 3 D (11.104) (11.105) (11.106) (11.107) (11.108) For example (np)(n0 d) gives also 6 dierent terms `1 = 1, (11.109) s1 = (11.110) L = S = ⇒ `2 = 2 1 s2 = 2 1, 2, 3 0, 1 1 P, 1 D, 1 F, 3 P, 3 D, 3 F For three electrons this process has to be extended stepwise. Physics of Atoms and Molecules (11.111) (11.112) (11.113) 11.6 Many electron atoms 163 Abbildung 11.5. Electronic conguration of the atoms and ionization energies. Physics of Atoms and Molecules 164 11 Two and many electron atoms C* Abbildung 11.6. Electronic conguration of the ground state of atoms with the spin orientations according to Hund's rules. C ∗ is the excited state of carbon leading to sp3 hybridization. Abbildung 11.7. Coupling of two equivalent p-electrons. The dotted lines correspond to energy levels which are forbidden by the Pauli exclusion principle. Physics of Atoms and Molecules 11.6 Many electron atoms 165 In the case of equivalent electrons in the same subshell the Pauli exclusion principle forbid two electrons to have the same quantum numbers. Also combinations of two electrons with quantum numbers are forbidden if they are identical upon electron exchange. For example (np)2 (Fig. 11.6)gives 3 dierent terms `1 = `2 = 1 1 s1 = s2 = 2 L = 0, 1, 2 S = 0, 1 ⇒ 1 S, 1 D, 3 P (11.114) (11.115) (11.116) (11.117) (11.118) The terms of 1 P , 3 S , and 3 D are not allowed due to Pauli exclusion principle. The forbidden 3 D state would have two electrons with the same spin quantum number in the same orbital m` = ±1. Furthermore, the total wavefunction, i.e. the product of the spatial wavefunction and spin wavefunction has to be antisymmetric. In the case of odd numbers for the total angular momentum (L = 1, 3, 5, ...) the spatial wavefunction is antisymmetric upon electron exchange. Singlet spin states have antisymmetric spin wavefunctions and tripett states symmetric spin functions upon electron exchange. Thus, the state 1 P would have a symmetric electron wavefunction and consequently is forbidden. Due to the Coulomb interaction and exchange force the levels split. The Hund's rules establish empirically the ground state conguration. 1 The term with the largest possible value of S for a given conguration has the lowest energy; the energy of the other terms increases with decreasing S. 2 For a given value of S , the term having the maximum possible value of L has the lowest energy. Explanation for the Hund's rules are the following (Fig. 11.6): If the spin quantum number is maximal, the spins are parallel and the electrons have to occupy dierent orbitals, reducing the Coulomb repulsion. In addition, parallel spins are favored because of the exchange force. For a given quantum number S maximal quantum numbers of L provide the lowest energies. This is because of the reduced Coulomb repulsion in orbitals with high L and therefore higher averaged electron-electron distance. Introducing spin-orbit coupling result in splitting of 2j + 1 degenerated levels. This is shown in Fig. (11.6) where the 3 P1 level split in three sublevels. Using the Physics of Atoms and Molecules 166 11 Two and many electron atoms P0 ml1=ml2=0 1S ml =+/-1 1 ml2=+/-1 1D ml1=0 ml =+/-1 2 3P biggest averaged electron-electron distance Abbildung 11.8. . quantum numbers of the hydrogen wave function `, s, j, mj we nd h X → − − → − → − → ξ(rk ) L k · S k i = Ah`sjmj | L i · S i |`sjmj i (11.119) k − → → − → − A h`sjmj | J 2 − L 2 − S 2 |`sjmj i 2 A = {j(j + 1) − `(` + 1) − s(s + 1)} 2 ⇒ E(j) − E(j − 1) = Aj = (11.120) (11.121) (11.122) Equation (11.122) is called Landé interval rule and is well satised experimentally if the atom is well described by the L-S coupling case (only coupling and weak). One nds A > 0: Open subshell is less than half full. The lowest value of j exhibits the lowest energy. A < 0: Open subshell is more than half full. The highest value of j exhibits the lowest energy. The strength of Spin-orbit coupling is proportional to Z 4 , so that weak spin-orbit coupling holds for Z ≤ 40. If spin-orbit coupling is dominant the total angular momentum couples for each Physics of Atoms and Molecules 11.6 Many electron atoms 167 Abbildung 11.9. The splitting of the ground state conguration of carbon due to spin-orbit coupling. individual electron i − → − → − → J i = L i + S i. (11.123) − → − → This is called j-j-coupling. Since closed shells have L = S = 0 only valence electrons are relevant. Pure j-j-coupling is rarely found and most atoms show intermediate coupling schemes. Physics of Atoms and Molecules 168 12 Molecules 12 Molecules Molecules consist of several atoms bound together. In addition to the atomic interactions, interactions between the electrons and nuclei of the atoms (for example A and B ) have to be taken into account. The forces resulting from electron-electron interaction and electron-nuclei interactions are comparable, but the masses of the nuclei M are much bigger than those of the electrons m m ≈ 10−3 . . . 10−5 . M (12.1) Therefore, the velocities of the nuclei are much smaller than that of the electrons and the kinetic energy of the nuclei is much smaller than that of the electrons. Thus, the electrons can adapt nearly instantaneously to the positions of the nuclei and the movement of electrons and nuclei can be separated. As long as this assumption holds we can use the product solution method to uncouple the movement of the electrons from the nuclei. Typical bond lengths of molecules H 2: e e r1 p r2 RA 0 RB MA p MB ~ 0.74 A Abbildung 12.1. Hydrogen molecule (H2 ) and coordinates of the two electrons and nucleic protons. The center of mass is indicated as 0. are in the range of Angstroms. The diatomic distance in the hydrogen and oxygen molecule (H2 , O2 ) was measured by neutron diraction to be ≈ 0.74Å and ≈ 1.21Å, respectively. In Fig. (12) the coordinates of the individual particles are Physics of Atoms and Molecules 12 Molecules 169 given. We can write them in the following form (here diatomic molecule) → − − → → − R = RB − RA (12.2) MA MB µ = (12.3) MA + MB → ~2 − → TN = − ∇ 2− (12.4) 2µ R N X →2 ~2 − → Te = − ∇− (12.5) 2m r i i=1 Ã N ! N 2 X X − → − e Z Z A B → V (R; → r 1, . . . , − r N) = − → − − − → → − − → 4πε0 i=1 | r i − R A | i=1 | r i − R B | + N X i,j=1 i>j ZA ZB e 2 e2 + → → 4πε0 R 4πε0 |− r i−− r j| (12.6) With this abbreviations the SE is given by − → − − → → → → [TN + Te + V ] ψ( R ; → r 1, . . . , − r N ) = Eψ( R ; − r 1, . . . , − r N ). (12.7) In order to get an impression of the energy scales we estimate the binding energy of the electrons by choosing their wavelengths of the order of the internuclear distance ~ R p2 ~2 ≈ ≈ 2m mR2 (12.8) p ≈ Ee Ev some eV s 1 k 1 = ~ω0 (v + ) = ~ (v + ). 2 µ 2 (12.9) (12.10) To dissociate a molecule one has to overcome the binding energy of the electrons or tear the atoms apart by the additional distance of R. With these assumptions we nd kR2 ≈ Ev ≈ (12.9) ≈ ≈ Ee , s H.O. s k Ee ≈~ µ µR2 s mEe2 ~ µ~2 r m Ee ≈ 10−2 × Ee . µ ~ Physics of Atoms and Molecules (12.11) (12.12) (12.13) (12.14) 170 12 Molecules Thus, the vibrational energies are about 0.1 eV and lie in the infrared spectral region. For example the vibrational energy of the diatomic molecule HCl is ν0 = ν0 1 ≈ 2990 cm−1 = . c0 λ[cm] (12.15) We estimate the rotational energy of a molecule by taking the Hamiltonian of a rigid rotor with the moment of inertia I , I = mR02 , that is − →2 − →2 LN N H= = (12.16) 2I 2I where m is the mass of the particle (point mass) xed at the distance R0 to the axis of rotation. The eigenfunctions are proportional to the spherical harmonics Y KMK (Θ, Φ) of the nucleic coordinates and the energy eigenvalues are Er = ~2 ~2 J(J + 1) = J(J + 1). 2I 2µR2 → − The operator N is the angular momentum operator of the nuclei with · µ ¶ ¸ → −2 1 ∂ ∂2 ∂ 1 2 N = −~ sin Θ + sin Θ ∂Θ ∂Θ sin2 Θ ∂Φ2 − → J(J + 1) Er = hψ| N 2 |ψi = ~2 2µR2 2 ~ m ≈ ≈ Ee 2µR2 µ ≈ 10−4 × Ee . (12.17) (12.18) (12.19) (12.20) (12.21) The rotational energies are in the range of 0.001 eV or in the far infrared / microwave spectral region 1 . . . 102 cm−1 . Therefore the contribution of the rotational energies are negligible in comparison to the electronic and vibrational contributions to the total energy. After estimation of the relative contributions to the total energy we see that the movements of the nuclei are much slower than that of the electrons. For every nucleic geometry the electrons have enough time to adopt to the actual situation and we solve the SE for the electrons for every xed and for → − the electrons static geometry of nuclei ( R xed). h i − → − − → → → − → − → → − → − → Te + V ( R ; r 1 , . . . , r N ) Φq ( R ; − r 1, . . . , → r N ) = Eq ( R )Φq ( R ; − r 1, . . . , − r N) − → → → − → − → − → He Φq ( R ; − r 1, . . . , → r N ) = Eq ( R )Φq ( R ; − r 1, . . . , − r N) → − He Φq = Eq ( R )Φq (12.22) He in equation (12.22) is called the electronic Hamiltonian of the electronic wave equation with Φq being the electronic wave functions. The electronic wave functi→ − ons and electronic energies depend on the parameter R for every state described Physics of Atoms and Molecules 12 Molecules 171 − → → by the quantum numbers q . The wave functions {Φq ( R ; − r )} form a complete and orthonormal set of functions: Z − → → − → → − − − → Φ∗s ( R ; − r 1, . . . , → r N )Φq ( R ; − r 1, . . . , → r N )d→ r 1 , · · · d− r N = δs,q . (12.23) − → → − Therefore, the exact total wave function ψ( R ; − r 1, . . . , → r N ) can be expanded for → − → − → every R with the help of the electronic wave functions {Φq ( R ; − r )} X − → → − → − → → − − ψ( R ; − r 1, . . . , → r N) = Fq ( R )Φq ( R ; − r 1, . . . , → r N) (12.24) q − → with the coecients Fq ( R ) representing the nuclear motion (vibration and rotation) when the electronic system is in the state q . Inserting equation (12.24) into (12.7) h iX X − → − − → − → → − → − → → → TN + Te + V ( R ; r ) Fq ( R )Φq ( R ; − r) = E Fq ( R )Φq ( R ; − r) q q (12.25) R − → and multiplying with Φ∗s . . . d→ r 1 , · · · d− r N we nd i XZ → − → h − → − → − → − → → → Φ∗s ( R ; − r ) TN + Te + V ( R ; → r ) − E Fq ( R )Φq ( R ; − r )d− r 1 , · · · d− rN = 0 q (12.26) 0 = X q = X q = X − → − → − hΦs |TN + Te + V ( R ; → r ) − E|Φq iFq ( R ) (12.27) → − hΦs |TN + He − E|Φq iFq ( R ) (12.28) → − → − hΦs |TN + Eq ( R ) − E|Φq iFq ( R ) (12.29) q = X q + 2 " µ ~ 1 ∂ ∂ R2 2 2µ R ∂R ∂R − → → − [Es ( R ) − E]Fs ( R ). hΦs | − ¶ − →2 # → − N − 2 2 |Φq iFq ( R ) ~R (12.30) − → → In equation (12.30) we used the orthonormality of {Φq ( R ; − r )} and the fact − → − → that Eq ( R ) only depends on R and E is eigenvalue of the total wave function − → → → ψ( R ; − r 1, . . . , − r N ). The set of coupled equations (12.30) is exact and equivalent to the full SE (12.7). The Born-Oppenheimer (BO) or adiabatic approximation Physics of Atoms and Molecules 172 12 Molecules will now be introduced. Since Φq varies slowly with respect to R, Θ and Φ the following relation holds ∂Φq ∂Fq | ¿ | | ∂R ∂R X X − → − → ⇒ hΦs |TN |Φq iFq ( R ) ≈ hΦs |Φq iTN Fq ( R ) | q (12.32) q    ~2 ∂ ∀ s: −  2µR2 ∂R (12.31) µ ∂ R ∂R 2 ¶ − →  → hΦs | N 2 |Φq i  Fs (− + +E (R) − E R ) = 0. (12.33) s  2 2µR | {z } ∼Erot Equation (12.33) is the nuclear wave equation. Since Erot is negligible small the nuclear wave equation simplies to · 2 ¸ −2 − → − → − → ~ → − ∇ R + Es ( R ) Fs ( R ) = E Fs ( R ), ∀s (12.34) 2µ In the BO approximation rst the electronic wave equation (12.22) with the wave − → → − → functions {Φq ( R ; − r )} are solved for parametric values of R to nd the energies − → − → Eq ( R ). With the electronic energies Eq ( R ) we can solve the nuclear wave equa- 0 q Potential E (R) (eV) 2 -2 -4 0 1 2 Nuclear distance R (Angstrom) Abbildung 12.2. Energy eigenvalue of the electronic wave function Eq plays the role of a potential for the nuclear wave function. Red horizontal lines indicate the energy eigenvalues of the nuclear vibrational energies. For dierent quantum numbers q the potentials are dierent. − → − → tion (12.33) to get the wave functions Fs ( R ), where the energy functions Es ( R ) play the role of a potential. The total wave functions ψs in the BO approximation has the product form − → − → − → − → − → ψs ( R ; → r 1, . . . , → r N ) = Fs ( R )Φs ( R ; − r 1, . . . , − r N ). (12.35) Physics of Atoms and Molecules 12.1 Diatomic molecules 173 − → − → − 2 For every nuclear coordinate R the electron probability density |Φq ( R ; → r )| is dierent and the equations (12.22)and (12.34) have to be solved. The nuclei move → − in the potential given by the eigenvalues of the electronic wave equation Es ( R ). 12.1 Diatomic molecules The electronic wave equation (12.22), in which only the Coulomb interactions are taken into account, is solved for diatomic molecules by using a system of coordinates OX , OY , OZ which is xed with respect to the molecule. The Z axis is chosen along the internuclear axis from B to A. The origin O is the center of mass and the spherical coordinates are R, Θ and Φ. This is called the body→ xed or molecular frame. The position vector − r i of an electron is the same in → the space-xed and body-xed frames, but the components of − r i are dierent in − → each frame. We will label the components of r i by xi , yi , and zi in the space-xed frame and by xi , y i , and z i in the body-xed frame. The eigenvalues Es (R) are independent of the frame of reference. xi = cos Θ cos Φ xi + cos Θ sin Φ yi − sin Θ zi y i = − sin Φ xi + cos Φ yi z i = sin Θ cos Φ xi + sin Θ sin Φ yi + cos Θ zi (12.36) (12.37) (12.38) The dierent frames of reference can be brought to coincidence by a rotation through an angle φ about the OZ axis and rotation about OY' axis through an angle Θ. We nd that ∂Φq i = − Ly Φq ∂Φ ·~ ¸ ∂Φq i i = − cos ΘLz + sin ΘLx Φq ∂Φ ~ ~ µ ¶ N X ∂ ∂ Lz = −i~ xi − yi ∂y ∂xi i i=1 (12.39) (12.40) (12.41) 12.2 Symmetry properties In the electronic wave equation the three components of the total electronic or− → bital angular momentum L commute with the Hamiltonian (11.90) of a many − → electron atom as well as L 2 . Thus, we have the good quantum numbers L and ML . The fact that L is approximately a good quantum number is used in the classication of atomic energy levels and terms (e.g. 3 P2 ). In contrast the internuclear axis of diatomic molecule picks out a single direction in space. If we choose this direction as the z axis, then the operator Lz commutes with He , but not Lx , − → Ly , and L 2 . Since the electronic molecular Hamiltonian He is invariant under Physics of Atoms and Molecules 174 12 Molecules rotations about the internuclear line or z axis, but not about x or y axis, only Lz commutes with the Hamiltonian. Therefore, simultaneous eigenfunctions of He and Lz can construct the electronic eigenfunctions Φs of a diatomic molecule and it is Lz = ML ~Φs , = ±Λ~Φs , ML = 0, ±1, ±2, . . . Λ = 0, 1, 2, . . . (12.42) (12.43) where Λ = |ML | is the projection of the total electronic orbital angular momentum on the internuclear axis. The azimuthal part of the Φs is therefore given by 1 Φs ∝ √ e±iΛΦ . (12.44) 2π By analogy with the spectroscopic notation S,P,D,F,... used for many electron atoms having the quantum number L the quantum number Λ is used for diatomic molecules Value of Λ 0 1 2 3 ... l l l l Code letter Σ Π ∆ Φ . . . (12.45) To describe transitions or the state characterized by individual electrons we use the quantum number λ = |m` | and the analogue notation Value of λ 0 1 2 3 ... l l l l Code letter σ π δ φ . . . (12.46) The electronic Hamiltonian H (without interaction to the nuclei) of a diatomic molecule is invariant under reections in all planes containing the internuclear line, for example the XZ plane. If this operation y i → −y i is described by the operator Ay then [Ay , He ] = 0 Lz = −i~ N µ X i=1 xi ∂ ∂ − yi ∂y i ∂xi (12.47) ¶ , Ay Lz = −Lz Ay N electrons (12.48) (12.49) If Λ 6= 0 (the Π, ∆, Φ . . . terms) the operator Ay acting on the wave function with eigenvalue Λ~ of Lz converts the wave function into another one with eigenvalue −Λ~. Both eigenfunctions have the same energy, because Ay commutes with He , and are doubly degenerate. If we consider the case Λ = 0 we have Σ states, that are non-degenerate and Physics of Atoms and Molecules 12.3 Neumann-Wigner non-crossing rule 175 simultaneous eigenfunction of He , Lz , and Ay can be constructed. The eigenvalues of Ay are ±1 and thus the Σ states of diatomic molecules are specied by Σ+ and Σ− for states left unchanged or changed upon reection in a plane containing the nuclei, respectively. In the special case of homonuclear diatomic molecules (such as H2 , O2 , N2 , etc.) there is an extra symmetry upon reecting the coordinates − − of all electrons with respect to the origin → r i → −→ r i , because of the center of symmetry of this molecule. This operator commutes with Lz and the electronic wave functions split into two sets, those that are even and do not change and those that are odd and change its sign. The even electronic wave functions are denoted by a subscript g called gerade states, the odd electronic wave functions denoted by a subscript u called ungerade states. For homonuclear diatomic molecules we have the following states − − + Σ+ g , Σu , Σg , Σu , Πg , Πu , . . . (12.50) → − − → Exchanging the nuclear coordinates R → − R is the same operation as rst rotating the molecule as a whole through 180◦ about the Y axis, followed by the → → operations y i → −y i and − r i → −− r i . If we denote the resultant spin of the − → → − individual electron spins by S with the usual eigenvalues of S 2 to be ~2 S(S + 1) we characterize the molecule by its multiplicity (2S + 1) as a left superscript. The term symbol is written 2S+1 3 Λ, ∆, for Λ = 0, 1, 2, . . . Λ = 2, S = 1 (12.51) (12.52) The ground state of a molecule is often labeled by the symbol X and since the ground state has very often maximum symmetry in diatomic molecules it is then denoted by X 1 Σ+ g . Excited states of the same multiplicity as the ground state are usually distinguished by the upper case letters A, B, C, . . . and those of dierent multiplicity by lower case letters a, b, c, . . .. Hyperne structure interactions due to coupling to the nuclear spins and the orbital motion and spin of the electrons are usually small and can be neglected. However, symmetry eects related to the spin of the nuclei have an important inuence on the structure of the wave functions especially if the molecule consist of identical nuclei. 12.3 Neumann-Wigner non-crossing rule For every electronic eigenfunction Φs of the molecule there is a multidimensional → − − → potential Es ( R ), which depends on the internuclear vector R . We want to know how these potential curves changes as R varies, and if is it possible that two Physics of Atoms and Molecules 176 12 Molecules potential curves intersect. If we suppose we have two dierent electronic potential curves E1 (R) and E2 (R) as depicted in Fig. 12.3 that are close but distinct at (0) (0) the internuclear distance Rc . The energies E1 and E2 are eigenvalues of the electronic Hamiltonian H0 = He (Rc ). The corresponding orthonormal electronic (0) (0) eigenfunctions are Φ1 and Φ2 . If we modify the internuclear distance Rc to E1(R) Es(R) E2(R) 0 Rc Rc+ ∆R R Abbildung 12.3. Angular momentum of a diatomic molecule with no coupling between the electronic and nuclear orbital motion and without spin coupling. Rc + ∆R with small ∆R, we can treat the new situation as a perturbation to the electronic Hamiltonian He (Rc + ∆R) = He (R) + H 0 = H0 + H 0 ∂H0 H 0 = ∆R ∂Rc (0) (0) Hij0 = hΦi |H 0 |Φj i, i, j = 1, 2. (12.53) (12.54) (12.55) (0) If we have a degenerate case with degeneracy of N and N wave functions ψku we have to solve the following determinant to obtain a non-trivial solution. (0) (0) (1) det|hψku |H 0 |ψks i − Ekr δus | = 0, (r, s, u = 1, 2, . . . , N ) (1) (1) (1) (12.56) Solving the determinant yields N real roots Ek1 , Ek2 , . . . , EkN . If all these roots are distinct the degeneracy is completely removed to rst order in the perturbation. Here, we have N = 2 and we have to solve the determinant at the position where the curves should cross, that is at Rc + ∆R and H0 + H 0 . It is (0) (0) (0) hΦi |H0 |Φj i = Ei δij , and we have the determinant ¯ ¯ ¯ ¯ (0) (0) (0) (0) hΦ1 |H 0 |Φ2 i ¯ ¯ hΦ1 |H0 + H 0 |Φ1 i − E (12.57) ¯ = 0. ¯ (0) (0) (0) (0) 0 0 ¯ hΦ2 |H |Φ1 i hΦ2 |H0 + H |Φ2 i − E ¯ Physics of Atoms and Molecules 12.4 Rotation and vibration of diatomic molecules 177 The potentials E1 (R) and E2 (R) are equal at position Rc +∆R if the two equations are simultaneously satised (0) (0) 0 0 E1 − E2 + H11 − H22 = 0, 0 H12 = 0. same energies for 1 and 2 (12.58) (12.59) The last term follows from solving the determinant with two identical energy values E , so that the potentials can cross. We have to distinguish two cases: (0) 0 The matrix element H12 is zero. That is the case if the two functions Φ1 (0) and Φ2 have dierent symmetries (for example having dierent values of Λ). Then it is possible to nd a certain value ∆R where the energies are the same E1 (Rc + ∆R) = E2 (Rc + ∆R). (0) (0) 0 If Φ1 and Φ2 have the same symmetry, then H12 will in general be nonzero. In almost all cases it is impossible to nd a single value ∆R for which the two conditions are satised simultaneously. Thus in general two electronic curves belonging to the same symmetry species cannot cross. This is called the non-crossing rule of von Neumann and Wigner. S S 1 0 25,5 Energy (h ) 20,5 15,5 10,5 5,5 0,5 -9 -6 -3 0 3 6 9 vibrational coordinate y Abbildung 12.4. Energy relaxation via internal conversion (IC). If two potentials have dif- ferent symmetries, the can cross. This is naturally the case for singlet and triplet potentials. However, if some coupling exist between two potentials S0 and S1 for example by vibrational coupling, the population of the level S1 can relax via internal conversion (IC) to the S0 state. In this case the IC increases with increasing overlap (and mixing due to coupling) of the vibrational wave functions in the two potentials. 12.4 Rotation and vibration of diatomic molecules In the nuclear wave equation (12.33)the rst term describes the kinetic energy due to radial motion (we are in the center of mass coordinate system) and the second Physics of Atoms and Molecules 178 12 Molecules term is the kinetic energy due to rotational motion. Es (R) acts as an eective → − potential. The orbital angular momentum N of the nuclei A and B (diatomic − → molecule) can be expressed by the total orbital angular momentum K , and the orbital angular momentum of the electrons → − − → − → N = K − L. (12.60) − → − → − → → − The orbital angular momentum is given by N = R × P , where P is the operator − → for the relative linear momentum of A and B and R the internuclear position → − − → vector. It follows that R · N = 0 and that the components of the electronic − → → − orbital angular momentum L and the total orbital angular momentum K along AB, which we dene as the z-axis are equal Lz = Kz . L N K A RA 0 RB B Abbildung 12.5. Angular momentum of a diatomic molecule with no coupling between the electronic and nuclear orbital motion and without spin coupling. − → − → Kz = K · R = Lz . (12.61) For an isolated molecule with negligible spin-dependent coupling and no coupling between electronic and nuclear orbital motion the wave function ψs must be a − → simultaneous eigenfunction of K 2 and Kz , where Kz is the component of K along the OZ (OA, OB ) space-xed axis. Thus we have that − →2 K ψs = K(K + 1)~2 ψs (12.62) where the rotational quantum number K is a positive integer or zero, and −K ≤ MK ≤ K : Kz ψs = MK ~ψs . (12.63) In addition equations (12.42, 12.43) show that ψs is an eigenfunction of Lz − → Lz ψs = Fs ( R )Lz Φs = ±~Λψs (12.64) Kz ψs = ±~Λψs (12.65) Physics of Atoms and Molecules 12.4 Rotation and vibration of diatomic molecules 179 and because of equation (12.61) ψs must also be an eigenfunction of Kz . The possible values of the quantum number K are (see Fig. 12.4) − → | K | ≥ Kz K = Λ, Λ + 1, Λ + 2, . . . (12.66) (12.67) Now we show the rotational kinetic energy term is given by − → − → − → − → − → − → − → 1 1 hΦs | N 2 |Φs iFs ( R ) = hΦs | K 2 + L 2 − 2 K · L |Φs iFs ( R ) 2 2 2µR 2µR − → − → → − → − 1 = hΦs | K 2 + L2z − 2 K · L + L2x + Ly2 |Φs iFs ( R ) 2 2µR − → ~ = [K(K + 1) − Λ2 ]Fs ( R ) 2 2µR → − 1 2 2 − hΦ |L L + 2N L |Φ iF ( R) + L + 2N s x x y y s s x y 2µR2 (12.68) Here we used the fact that Φs are orthonormal eigenfunctions and eigenstates − → of L z . Because of the axial symmetry around the z axis the expectation values hLy i, and hLx i are zero. The relative notion of nuclei can now be expressed as · µ ¶ ¸ → − − → ~2 1 ∂ K(K + 1) 2 ∂ 0 = − R − Fs ( R ) + [Es0 (R) − E]Fs ( R ) 2 2 2µ R ∂R ∂R R (12.69) 2 2 1 Λ~ − hΦs |L2x + Ly2 |Φs i Es0 (R) = Es (R) − 2 2µR 2µR2 (12.70) → − The energy Es0 (R) acts as an eective potential for the wave function Fs ( R ) and depends only on the electronic states. Since the last two terms are much smaller than the rst one, due to the greater mass of µ with respect to m we use the approximation (12.71) Es0 (R) = Es (R). − → The wave function for the nuclear motion Fs ( R ) can be expressed in terms of the product of a radial function FsvK (R) and an angular function Y ΛKMK (Θ, Φ) − → 1 Fs ( R ) = FsvK (R)Y ΛKMK (Θ, Φ). R (12.72) The quantum number v is called vibrational quantum number and assigns dierent solutions of equation (12.69) and is related to molecular vibrations. For a given electronic state s the dierent states called rovibronic states are labeled Physics of Atoms and Molecules 180 12 Molecules by the rotational quantum number The angular function Y ΛKMK (Θ, Φ) − → Λ. For Σ states (Λ = 0) it is K = spherical harmonics YKMK (Θ, Φ). K and the vibrational quantum number v . − → are eigenfunctions of K 2 and Kz for a given − → N and the angular functions reduce to the Y ΛKMK (Θ, Φ) = NΛKMK × eiMK Φ (1 − cos Θ)(Λ−MK )/2 (1 + cos Θ)−(Λ+MK )/2 # "µ ¶(K−MK ) © ª ∂ K−Λ K+Λ (1 − cos Θ) (1 + cos Θ) ) (12.73) × ∂ cos Θ s 1 (2K + 1)(k + MK )! NΛKMK = (−1)K √ (12.74) 2K+1 (K + Λ)!(K − Λ)!(K − MK )! 2π 2 Inserting equation (12.72) into equation (12.69) and dividing by the angular functions we obtain the radial equation for the nuclei ¶ ¸ · 2µ 2 d K(K + 1) ~ − + Es (R) − EsvK FsvK (R) (12.75) 0 = − 2µ dR2 R2 ¯ ¯ 2 ¯ dEs ¯¯ 1 2 d Es ¯ Es (R) = Es (R0 )+ (R − R0 ) (R − R ) + + ... 0 dR ¯R=R0 2 dR2 ¯R=R0 (12.76) Since Es (R) has a minimum at R = R0 the second term on the right hand side of equation (12.76) vanishes. Neglecting the third and higher order terms we get the parabolic approximation that correspond to the harmonic oscillator 1 Es (R) = Es (R0 ) + ks (R − R0 )2 ¯ 2 1 d2 Es ¯¯ ks = . 2 dR2 ¯R=R0 (12.77) (12.78) Inserting equation (12.77) into (12.75) we nd an eigenvalue equation of the vibrational energy Ev · 2 2 ¸ ~ d 1 2 − + ks (R − R0 ) − Ev ψv = 0. (12.79) 2µ dR2 2 This is the same eigenvalue equation than for the harmonic oscillator and we nd the same eigenfunctions as in the case of the harmonic oscillator and eigenvalues given by µ ¶ µ ¶ 1 1 Ev = ~ω0 v + = hν0 v + , v = 0, 1, 2, . . . (12.80) 2 2 with ω0 = (ks /µ)1/2 . Additionally, we can simplify equation (12.75) by approximating the internuclear distance R by R0 , that result in an expression for the Physics of Atoms and Molecules 12.4 Rotation and vibration of diatomic molecules 181 rotational energy Er ~2 ~2 Er = K(K +1) = K(K +1) ≡ BK(K +1), 2µR02 2I0 K = Λ, Λ+1, . . . (12.81) I0 is the moment of inertia of the molecule for an equilibrium distance R0 and reduced mass µ. The constant B describing molecular properties is called rotational constant. The equilibrium distance and hence B depends on the electronic state s. The total energy EsvK is the sum of the individual energies (12.82) EsvK = Es (R0 ) + Ev + Er . For allowed pure rotational transitions the permanent electric dipole moment has to be non-zero. In the dipole approximation the selection rules are for pure rotational transitions (12.83) (12.84) ∆K = ±1 ∆MK = 0, ±1. Note, the energy of the molecule cannot depend on MK in the absence of external elds and with no preferred direction in space. The energy is twofold degenerated for Λ 6= 0 with degeneracy 2(2K + 1) and for Σ states with Λ = 0 we nd degeneracy of the energy levels of 2K + 1. The levels split in the presence of an electric eld (Stark eect). In the representation of the principal axis system of a rigid body the kinetic energy of a rotating body is Er = 1 2 1 2 1 2 Ka + Kb + K 2Ia 2Ib 2Ic c → −2 K = Ka2 + Kb2 + Kc2 (12.85) (12.86) where Ia , Ib , and Ic are the principal moments of inertia about axis a, b, and c with its components of the angular momentum Ka , Kb , and Kc about the axis. Diatomic molecules are axially symmetric about the internuclear line, with Ia = Ib , Ic = Iz , and Kc = Kz . The system is called symmetrical top molecule 1 2 1 (Ka2 + Kb2 ) + K 2Ia 2I c µ c ¶ 1 1 1 = K(K + 1) + − Λ2 ~2 , 2Ia 2Ic 2Ia Er = Er (12.87) K = Λ, Λ + 1, . . . (12.88) The rotational energy in equation (12.88) consist of two terms. The rst term has a rotational axis that goes through the center of mass with Ia = µR02 and depend Physics of Atoms and Molecules 182 12 Molecules on K(K + 1). The second term depends on the moment of inertia 'Ic ' and on the eigenvalue of the electronic state Λ2 ~2 . If we include the second term in the electronic energy Es (R0 ) we get the same description of the rotational energy as in equation (12.81). The approximation of the harmonic oscillator potential is only accurate in the vicinity of the equilibrium internuclear distance R0 . Using this potential we immediately nd the eigenfunctions, which is more dicult for real potentials. In the harmonic approximation one can identify 3N − 5 normal vibrations for linear molecules and 3N − 6 normal vibrations for non-linear molecules. N denotes the number of atoms (see Fig. 12.4). For a linear molecule like carbon dioxide the potential for small oscillations around the equilibrium position by an asymmetric stretching will be V = 21 k(∆x1 − ∆x2 )2 + 12 k(∆x2 − ∆x3 )2 , with xi are the coordinates for the three atoms along the internuclear axis. Solving the linear equations of the atomic coordinates result in normal coordinates describing normal vibrations. The HO potential underestimates the real potential for smaller R and is Normal vibrations: CO2 O 1337 cm -1 2349 cm -1 C H2O O H O ν2 symm. stretch ν3 H 3657 cm-1 3756 cm -1 asymm. stretch ν1 bending vib. 667 cm-1 2x 1595 cm-1 Number of normal vibrations: Linear molecules: 3N-5 Non-linear molecules: 3N-6 Abbildung 12.6. Normal modes of carbon dioxide and water. far to strong for large internuclear distances R. A much better description of the electronic potential for the nuclear vibration is given by (see Fig. 12.4) £ ¤ Es (R) = Es (∞) + VM (R) = Es (∞) + De e−2α(R−R0 ) − 2e−α(R−R0 )(12.89) VM (R) = De [−1 + α2 (R − R0 )2 + . . .]. (12.90) Physics of Atoms and Molecules 12.4 Rotation and vibration of diatomic molecules 183 0 q Potential E (R) (eV) 2 -2 -4 0 1 2 Nuclear distance R (Angstrom) Abbildung 12.7. Comparison between the harmonic oscillator potential (blue) and the much more realistic Morse potential. The terms R0 , De , and α are positive constants and are properties of the molecules. Upon expanding VM (R) we nd by comparison of the coecients r 1 De α = ks ; 2 2 α = ω0 µ 2De (12.91) and the energies "µ Ev = ~ω0 1 v+ 2 ¶ µ 1 −β v+ 2 ~ω02 4De β = χe = ¿ 1 (12.92) (12.93) βω0 = D0 = De − ¶2 # (12.94) ~ω0 . 2 (12.95) The quantity βω0 is known as the anharmonicity constant and the dissociation energy of the molecule (from the vibrational ground state at T = 0K) is given by D0 . A deciency of the Morse potential is that it is nite at R = 0, in contrast to the true interaction V (R) which is innite at the origin. Another defect is that the potential falls o exponentially at large R instead of showing the R−6 fall-o due to van der Waals interaction between two neutral atoms which are far apart. The wave functions of the Morse potential Ψj,v are given by Physics of Atoms and Molecules 184 12 Molecules Molecule R0 (Å) D0 (eV) ν 0→1 (cm−1 ) B/(hc) (cm−1 ) 1030 |D| (Cm) H2 O2 Cl2 N2 CO NO HCl N aCl 0.742 1.21 1.99 1.09 1.13 1.15 1.28 2.36 4.5 5.15 2.51 9.83 11.13 6.5 4.46 4.24 4159.2 1556.3 565.9 2330.7 2143.3 1876 2885.6 378 60.8 1.45 0.244 2.01 1.93 1.70 10.6 0.190 0.40 0.50 3.53 28.1 Tabelle 12.1. Equilibrium distance R0 , dissociation energy D0 , fundamental vibrational fre- quency ν 0 = ν0 /c, rotational constant B and the magnitude of the electric dipole moment in debyes (1 debye = 3.36 × 10−30 Cm). z = ke−α[(R−R0 )] r 8µDe k = 2 2 pα ~ v!(k − 2v − 1)Γ(k − v) Av = k−2s−1 v X √ (−1)s z 2 −z/2 Ψv = αAv e s!(v − s)!Γ(k − v − s) s=0 vM ax = 1 1 − . 2β 2 1 eV = 8065.48 cm−1 = 1.60219 × 10−19 J 1 cm−1 = 1.98648 × 10−23 J Physics of Atoms and Molecules (12.96) (12.97) (12.98) (12.99) (12.100) (12.101) (12.102) (12.103) 12.4 Rotation and vibration of diatomic molecules 185 0,5 Energy (eV) 0,0 -0,5 -1,0 Morse potential; D=-2.0 eV v=0 (E=0.0726358 eV) v=4 (E=0.605357 eV) -1,5 v =12 (E=1.41285 eV) v=26 (E=1.99836 eV) -2,0 1 2 3 4 5 6 7 Displacement Abbildung 12.8. Morse Potential with some selected eigenfunctions of the Morse potential. 1,5 v = 0 v = 1 1,0 Wave function v = 2 v = 3 0,5 v = 4 v = 5 0,0 -0,5 -1,0 0,5 1,0 1,5 2,0 Displacement Abbildung 12.9. Asymmetric shift of the probability of the eigenfunctions of the Morse potential to larger nuclear distances (displacement) upon going from v = 0 to v = 5. Physics of Atoms and Molecules 186 12 Molecules Amplitude 0,5 0,0 v = 23 v = 24 v = 25 -0,5 v = 26 0 5 10 15 20 25 30 Displacement Abbildung 12.10. Enhancement of the wave function amplitude at larger displacement for high vibrational excitations near the dissociation limit. Abbildung 12.11. Explicitly calculated dependence of the N-H stretching vibration of a N H3 N H4+ complex with the hydrogen bonding stretching. Physics of Atoms and Molecules 12.4 Rotation and vibration of diatomic molecules 187 12.4.1 Centrifugal distortion or rotation-vibration coupling The rotational energy Er was approximated by the simple 'rigid rotator' expression with a xed internuclear distance at the equilibrium distance R0 leading to a decoupling of vibrational and rotational motions. Without this approximation we get a rened equation ¸ · 2 2 ~ d (12.104) 0 = − + Vef f (R) − EsvK FsvK (R) 2µ dR2 ~2 K(K + 1) Vef f (R) = VM (R) + , K = Λ, Λ + 1, . . . (12.105) 2µ R2 1 Vef f (R) = V0 + k s (R − R1 )2 + c1 (R − R1 )3 + c2 (R − R1 )4 (12.106) 2 In equation (12.106) we expanded Vef f (R) to the fourth order about its minimum V0 at R = R1 . Due to the fact that R is no longer a constant with R = R0 we nd a new equilibrium position at R1 . The new equilibrium position R1 coincides only with R0 if K = 0. If we use the simple approximation of c1 = c2 = 0 and k s = ks in equation (??) we nd the new equilibrium position to be ~2 K(K + 1) R1 ∼ . (12.107) = R0 + 2µ α2 R03 De The new energy eigenvalues can be calculated by treating the coecients c1 and c2 as perturbations. The energy eigenvalues are given to second order by "µ ¶ µ ¶2 # 1 1 ~2 EsvK = −De + ~ω0 v+ −β v+ + K(K + 1) 2 2 2µR02 µ ¶ 1 bK 2 (K + 1)2 (12.108) − a v+ K(K + 1) − | {z } 2 | {z } stretching correction to the rigid rotator a = b = rotation−vibration coupling µ ¶ 3~3 ω0 1 1− 4µαR03 De αR0 4 ~ , 4µ2 α2 R06 De stretching constant. (12.109) (12.110) In contrast to molecular vibrations in the condensed phase (liquid, solid phase) excitations of molecular vibrations in the gas phase are coupled to rotational transitions. If we assume excitation from the vibrational levels v → v 0 and rotational levels K → K 0 and no rotation-vibration coupling we nd the following energy dierences: µ µ ¶ ¶ 1 1 0 0 0 0 + B K (K + 1) − hcν v + − BK(K + 1) E2 − E1 = hcν v + 2 2 (12.111) K 0 =K−1 ⇒ ∆E = hcν − K(B 0 + B) + K 2 (B 0 − B) Physics of Atoms and Molecules (12.112) 188 12 Molecules The rotational constants B and B 0 are distinct for the dierent vibrational quantum numbers v and v 0 . From the measurement of the spectral positions of the absorption peaks one can determine the changes of the rotational constants and thus the averaged internuclear distance for simple molecules. In the case of the linear triatomic 12 CO2 molecule the internuclear distance between the oxygen and carbon atom is given by RC=O = 1.1588 × 10−8 cm. The total wave function ψ is given by Ψ = ψχN = Φs χSMs 1 FsvK (R)Y ΛKMK (Θ, Φ)χN . R (12.113) In equation (12.113) Φs is the spatial part of the electronic wave function, χSMs is R-branch P-branch Resolution: P-branch Energy in cm-1 Abbildung 12.12. Rotation-Vibration spectrum of 12 CO2 and 13 CO2 . There are two rotational transition branches R (J → J + 1) and P (J → J − 1) coupled to the vibrational transition v = 0 → 1. The dip in the middle of the two branches is related to the energy dierence ∆ν = ν 1 − ν 0 = 2349 cm−1 of a vibrational transition of 12 CO2 . The analogue energy for 13 CO2 has a value of 2282 cm−1 . the spin function of the electrons, and FsvK (R) and Y ΛKMK (Θ, Φ) are the nuclear vibrational and rotational wave functions, respectively. The function χN is the nuclear spin function. The function Φs is even or odd under the transformation − → − → + − R → − R . For example for Λ = 0 the Σ+ g and Σu states are even while the Σu − → − → and Σ− g states are odd. The function χSMs is unaected upon R → − R . The vibrational wave function FsvK (R) depends only on the magnitude R of the vector − → − → − → R and is unaltered when R → − R . The rotational wave function Y ΛKMK (Θ, Φ) Physics of Atoms and Molecules 12.4 Rotation and vibration of diatomic molecules 189 Energy in (cm-1) P-branch Rotational quantum number K Abbildung 12.13. Spectral positions of the rotation-vibration absorption maxima as a function of K. The solid line is a polynomial t to second order with coecients: A0 = 2349.71 ± 0.04, A1 = −0.781 ± 0.004, A2 = −0.00293 ± 0.00007. is either even (for K = 0, 2, 4, . . .) or odd (for K = 1, 3, 4, . . .) under the → − − → transformation R → − R . The overall symmetrical (S) or antisymmetrical (A) character of the function ψ is given as follows: g + g − u + u− K even S A A S K odd A S S A (12.114) The nuclear spin function χN can be symmetric or antisymmetric with respect to the interchange of the nuclei. For example 16 O2 : 16 O2 has two spinless nuclei; IN = 0 χN is a constant and trivially symmetric in the interchange of the nuclei. The total wave function Ψ has to be symmetric (Boson) Ψ,χN , ψ have to be symmetric If the electronic state is even (g + , u− ), only even K are allowed and if the electronic state is odd (g − , u+ ), only odd K are allowed The electronic ground state of O2 is a 3 Σ− g state Only odd rotational quantum numbers K are allowed. The ground state of the 16 O2 molecule is given by K = 1. Physics of Atoms and Molecules 190 12 Molecules The same eect is observed in CO2 which has a ground state 1 Σ+ g and therefore only even rotational quantum numbers K are allowed (∆K = ±2). This results in reduced numbers of vibration-rotation transitions. If the symmetry is reduced 16 O = C =18 O all vibration-rotation transitions can be observed. The degeneracy of rotational levels is (2K + 1) (for Σ states), and the relative population ratios are given by NK gK −(EK −E0 )/(kB T ) = e = (2K + 1)e−BhcK(K+1)/(kB T ) . N0 g0 (12.115) Since the rotational energies are small compared to the room temperature kB TT =293K ≈ 1 eV ≈ 204cm−1 40 (12.116) we can expect population of higher rotational levels at room temperature. The rotational level with the highest population Kmax is given by r Kmax = 1 kB T − . 2hcB 2 (12.117) That the population varies for dierent rotational quantum numbers K is visible in Fig. 12.4.1. In this gure the rovibronic absorption spectrum of the triatomic linear molecule 12 CO2 is presented. The vibration that is excited is the asymmetric stretching vibration which is infrared active and Raman inactive. Changing the vibrational quantum number vas of the asymmetric stretching vibration is always connected with changing the rotational quantum number K . This is due to the fact that changing the average bond length by going to higher or lower vibrational quantum numbers changes the moment of inertia and therefore the rotational quantum number (∆K ± 1). Excitation of the bending vibrations of 12 CO2 changes only the rotational quantum number for rotations with rotation axis perpendicular to the bending vibration plane (∆K ±1). Rotations parallel to the bending vibration plane do not change K and lead to the so called Q-branch with ∆K = 0. In general the expectation value of the transition dipole moment (dipole appro- Physics of Atoms and Molecules 12.4 Rotation and vibration of diatomic molecules ximation) is given by → h− µ Ti ZZ = Y − − ∗ → ∗ → ΛK 0 MK 0 Φf ( R , r )Ff v 0 K 0 (R) 191 " −e N X i=1 × = i=f, v=v 0 = − → → Y ΛKMK Φi ( R , − r )FivK (R)d cos ΘdΦ ZZ ∗ → Y ΛK 0 MK 0 − µ i→f,v→v0 Y ΛKMK d cos ΘdΦ ZZ ∗ → Y ΛK 0 MK 0 − µ perm Y ΛKMK d cos ΘdΦ. − → r i+e M X − → Zj R j # j=1 (12.118) (12.119) (12.120) → In equation 12.120 the term − µ perm is the permanent dipole moment of a molecule. HCl has a permanent dipole moment in contrast to homonuclear diatomic molecules such as H2 or N2 . Molecules with zero permanent dipole moment have no infrared active rotational transitions. For molecules with non vanishing permanent dipole moment the selection rules for pure rotational transitions are in analogy to the electronically transitions of atoms (dipole approximation): ∆K = ±1 ∆MK = 0, ±1 (12.121) (12.122) 12.4.2 Electronic spin and Hund's cases The most important magnetic interaction in molecules is the coupling of electronic − → − → spin S and the electronic orbital angular momentum L . If the nuclear spins are − → ignored, the total angular momentum operator J for a diatomic molecule is given by → − − → → − − → → − − → J = L +N + S =K+ S (12.123) → − where N is in a direction orthogonal to the internuclear line. In the case of |∆E| À |A| À B that is the electrostatic interaction is much larger than the spinorbit interaction which is in turn larger than the rotational energy. An example is given by the A2 Π term of CO+ . In the electronic ground state X 2 Σ+ it is found that |∆E| = 20733 cm−1 , the spin-orbit constant is A = −117 cm−1 and the rotational constant is B = 1.6 cm−1 . The electrostatic interaction has − → axial symmetry and this causes L to precess about the internuclear axis, and therefore Lz = ±Λ~ and hLx i = hLx i = 0. Since in this case the rotation of the molecule is slow, and the spin-orbit interaction is large compared with the − → rotational energy, the spin angular momentum S will also tend to precess about − → the internuclear axis. Thus, simultaneous eigenfunctions of S 2 and Sz can be found with eigenvalues of S(S + 1)~2 and Σ~, respectively. Since Nz = 0 it is Jz = L z + S z Ω = ±Λ + Σ Physics of Atoms and Molecules (12.124) (12.125) 192 12 Molecules where Ω~ are the eigenvalues of Jz and the quantum number Σ can take the (2S + 1) values −S, −S + 1, . . . , S . The quantum numbers Λ and Σ are the − → − → projections of L and S on the internuclear line. The additional L-S coupling leads to an energy given by ∆E = AΛΣ. (12.126) For a given number Λ the energy levels split into (2S + 1) multiplet components, which are equally space in energy by the factor A. For example if Λ = 1 and − → S = 1, we have Ω = 0, 1, 2 and 3 Π0 , 3 Π1 , and 3 Π2 . Since N 2 is now given by − → − → − → − → → − N = J − Ω~ R = J − Ω (12.127) and we can rewrite the rotational energy by Ω2 |{z} Er0 = B[J(J + 1) − ] (12.128) depends on the electronic state Er = B[J(J + 1)], J = Ω, Ω + 1, . . . (12.129) → − Since | J | ≥ Jz and J ≥ Ω this can be used to determine an unknown quantum number Ω by missing spectral lines with lower values of J . 12.5 LCAO: Linear combination of atomic orbitals To determine the orbitals of molecules we start with linear combinations of single electron atomic orbitals. This approach simplies wave function optimization with variational methods (Rayleigh-Ritz). We dene a test function ψtest X ψtest ≡ ψ = cj ϕj (12.130) j constructed by a complete basis set of single electron atomic orbitals {ϕi }. We have to minimize RP ∗ ∗ P R ∗ ci ϕi H cj ϕj dV ψ HψdV i j ² = R ∗ = R P ∗ ∗P (12.131) ψ ψdV ci ϕi cj ϕj dV i j PP ∗ ci cj Hij i j = PP ∗ (12.132) ci cj Sij i j with the following integrals Z Hij = ϕ∗i Hϕj dV Z Sij = ϕ∗i ϕj dV, (12.133) overlap integrals for i 6= j. Physics of Atoms and Molecules (12.134) 12.5 LCAO: Linear combination of atomic orbitals 193 To minimize (or maximize) ² we have to full the equations ∀i ∂² ∂² = ∗ =0 ∂ci ∂ci (12.135) Using equation (??) together with equation (12.132) we have XX i X XX j ∂ , ∂² =0 ∂c∗ ∂c∗ i i ⇒ c∗i cj Hij = ² i X Hij cj = ² X j c∗i cj Sij (12.136) j (12.137) Sij cj j (Hij − ²Sij ) cj = 0 (12.138) det|Hij − ²Sij | = 0 (12.139) j Equation (12.138) is a homogeneous system of linear equations (for all i's). We have to solve it in a non-trivial way to determine the coecients {cj } and build the optimized eigenfunction of the molecular orbital. Thus, the determinant in equation (12.139) has to vanish. As a result we obtain solutions of ²α , which have to be inserted into equation (12.138) to determine the coecients {cαj }. In the case of the H2+ molecule we choose the normalized 1s atomic orbitals (AO) ϕ1 and ϕ2 which are centered around the protons A and B (see Fig. 12.5). r 1 −rA /a0 e a0 π r 1 −rB /a0 = ϕ1s (B) = ϕ2 (rB ) = e a0 π → − − → − → = |→ r − R A |; rB = |− r − R B| → − − → − → → = → r − R /2; − rB=− r + R /2 ϕ1 = ϕ1s (A) = ϕ1 (rA ) = (12.140) ϕ2 (12.141) rA − → rA (12.142) (12.143) Since ϕ1 and ϕ2 are normalized we have S11 = S22 = 1, we nd the overlap integral S(R) to be Z S(R) = S12 (R) = Ã → r ϕ∗1 (rA )ϕ2 (rB )d− = R 1 + 1+ a0 3 µ R a0 ¶2 ! e−R/a0 (12.144) The overlap integral is a two center integral and can be solved using confocal elliptical coordinates. The matrix elements Hij are determined by using the fact Physics of Atoms and Molecules 194 12 Molecules + H 2: rA rB r O Abbildung 12.14. Coordinate system for the molecular ion H2+ . ϕ1 R A B ϕ2 R A B Abbildung 12.15. . that ϕ1 and ϕ2 are eigenfunctions of the hydrogen atoms A and B. e2 4πε0 R →2 ~2 − → − ∇− 2m r B Hel = HA + HB + →2 ~2 − → ∇− 2m r A e2 e2 e2 − − + → − − → − → 4πε0 |→ r − R A | 4πε0 |− r − R B | 4πε0 R = − Physics of Atoms and Molecules (12.145) (12.146) 12.5 LCAO: Linear combination of atomic orbitals 195 We can directly evaluate the matrix elements Z → H11 = ϕ∗1 (rA )Hel ϕ1 (rA )d− r (12.147) Z Z e2 e2 → − → ∗ = ϕ1 (rA )HA ϕ1 (rA )d r − ϕ∗1 (rA ) ϕ1 (rA )d− r + 4πε0 rB 4πε0 R Z 2 2 e e → = E1s + (12.148) − ϕ∗1 (rA ) ϕ1 (rA )d− r 4πε0 R 4πε0 rB {z } | Coulomb integral J 2 = E1s + e − J. 4πε0 R (12.149) The integral J in equation (12.149) describes the electrostatic attraction of −e|ϕ1 |2 from the nucleus A nucleic charge of the atom B . The CouR on the → lomb integral is positive (( e2 |ϕ1 |2 d− r )). Since the system is symmetric under exchange of nuclei A and B the matrix elements H11 = H22 are identical. The overlap integrals are given by Z → H12 = ϕ∗1 (rA )Hel ϕ2 (rB )d− r (12.150) · Z i e2 e2 → ∗ r (12.151) = ϕ1 (rA ) − + + HB ϕ2 (rB ) d− 4πε0 rA 4πε0 R | {z } µ = 2 E1s + µ = e 4πε0 R 2 E1s + e 4πε0 R ¶ E1s ϕ2 (rB ) Z ϕ∗1 (rA ) S− | e2 → ϕ2 (rB )d− r 4πε0 rA {z } (12.152) overlap integral K ¶ S(R) − K. (12.153) The overlap integral K describes the electronic overlap probability density −eϕ∗1 (rA )ϕ2 (rB ) which interacts with the nucleic charge of atom A. Due to symmetry reasons it has to be H12 = H21 . The term K in equation (12.153) is without classical interpretation, but represents the term that is responsible for the binding of dierent atoms. Now we have to solve the determinant: ¯ ¯ ¯ H11 − ² H12 − ²S ¯ ¯ ¯ (12.154) ¯ H21 − ²S H22 − ² ¯ = 0. This leads to a quadratic equation with solutions for the energies H11 + H12 1+S H11 − H12 = . 1−S ²+ = (12.155) ²− (12.156) Physics of Atoms and Molecules 196 12 Molecules Inserting the solutions into the system of equations we have the coecients 1 = c+ c+ 1 2 = p c− 1 2(1 + S(R)) 1 = −c− 2 = p 2(1 − S(R)) (12.157) (12.158) and can build the eigenfunctions ψ+ and ψ− 1 ψ+ = p 2(1 + S(R)) 1 ψ− = p 2(1 − S(R)) [ϕ1s (A) + ϕ1s (B)] (12.159) [ϕ1s (A) − ϕ1s (B)] (12.160) (12.161) The eigenfunctions ψ+ and ψ− given in equations (12.159) and (12.160), respec− − tively, are gerade and ungerade functions under the inversion operation → r → −→ r and represent the molecular orbitals for H2+ . The wave function ψ+ is the binding wave function and exhibit an increased electron density between the two nuclei A and B (see Fig. 12.5), whereas the wave function ψ− exhibits a knot at the center of inversion symmetry and a strongly reduced electron density between the two nuclei. The electron density is given by −e [ϕ∗ (A) ± ϕ∗1s (B)][ϕ1s (A) ± ϕ1s (B)] (12.162) 2(1 ± S) 1s −e [ϕ∗ (A)ϕ1s (B) + ϕ∗1s (B)ϕ1s (A)] = [|ϕ1s (A)|2 + |ϕ1s (B)|2 ] ± 1s 2(1 ± S) 2(1 ± S) (12.163) e|ψ± |2 = Physics of Atoms and Molecules 12.5 LCAO: Linear combination of atomic orbitals 197 Abbildung 12.16. Orbital wave functions of the hydrogen ion. The wave functions ψ+ and ψ− represent the binding and antibinding orbitals of the hydrogen ion. The binding orbital exhibits increased electron density between the nuclei, while the antibinding orbital shows reduced electron density between the nuclei. σ+ σ∗- Abbildung 12.17. The binding and antibinding molecular orbitals of the hydrogen ion has axial symmetry. The two energies of the molecular orbitals are given by H11 + H12 ²+ (R) = 1 + S(R) ³ ´ 2 2 E1s + 4πεe 0 R − J(R) + E1s + 4πεe 0 R S(R) − K(R) = 1 + S(R) 2 J(R) + K(R) e − = E1s + 4πε0 R 1 + S(R) H11 − H12 e2 J(R) − K(R) ²− (R) = = E1s + − 1 − S(R) 4πε0 R 1 − S(R) ¾ ½ 2 R R + 2 e−R/a0 S(R) = 1+ a0 3a0 ¶ ½ µ ¾ 2 e R −2R/a0 J(R) = e 1− 1+ 4πε0 R a0 ½ ¾ R e2 1+ e−R/a0 . K(R) = 4πε0 a0 a0 Physics of Atoms and Molecules (12.164) (12.165) (12.166) (12.167) (12.168) (12.169) (12.170) 198 12 Molecules The integrals J(R), K(R), and S(R) were solved by the use of confocal elliptical coordinates. The molecular ground state is connected with the wave function ψ+ (binding state). To compare these energies with the dissociation energy De one ε-, Ψ-, 1σu E1s, Ψ1s(A) E1s, Ψ1s(B) ε+, Ψ+, 1σg Abbildung 12.18. Energy splitting of the molecular orbital states. has to calculate the full dependence of ²+ (R) of the internuclear distance R. The equilibrium distance R0 is calculated to be R0 = 1.32 Åand the dissociation energy is calculated to be De = E1s − ²+ = 1.77 eV. The experimental values are R0 = 1.06 Åand De = 2.79 eV. The discrepancy is due to the fact that for R = 0 the charge at the origin is 2, comparable to the helium atom. But in the case of the helium atom we saw, that the eective charge is less than 2. The next step ∗ is to optimize the molecular orbitals with varied nuclear charge (e−Z r/a0 ). Up to now we considered only homonuclear diatomic molecules where the covalent bonding is strong. In general H11 6= H22 and if we set S(R) = 0 we nd for heteronuclear diatomic molecules: 0 = (H11 − ²)(H22 − ²) − H12 H21 ∗ H21 = H12 ; |H12 |2 = A2 > 0 0 = (H11 − ²)(H22 − ²) − |H12 |2 sµ ¶2 H11 + H22 H11 − H22 ± + A2 ²± = 2 2 s H11 + H22 H11 − H22 4A2 = ± 1+ 2 2 (H11 − H22 )2 (12.171) (12.172) (12.173) (12.174) (12.175) For many heteronuclear molecules it is |H11 − H22 | À A so that we can expand the square root and get A2 H11 − H22 A2 = H22 − H11 − H22 ²− = H11 + (12.176) ²+ (12.177) The lowering of the lowest energy level is now given by the term Physics of Atoms and Molecules 12.5 LCAO: Linear combination of atomic orbitals 199 Abbildung 12.19. The lowest electronic energy curves of H2+ . The curves labeled A show the calculated wave functions using LCAO, while the curves labeled B show the exact solutions. µ ¶ ∗ A2 H12 ∝ H12 ¿ H12 H11 − H22 H11 − H22 µ ¶ c1 H12 H11 − H22 ∝ = À1 c2 − ²− − H11 H12 µ ¶ c1 H12 H12 ∝ ∼ ¿1 c2 + ²+ − H11 H22 − H11 (12.178) (12.179) (12.180) In the case of heteronuclear bindings the binding is strong if the interacting atomic orbitals are similar and H11 ≈ H22 . If this is not the case, the ratios of coecients for ψ+ and ψ− is very dierent, meaning the molecular orbitals are poorly mixed (ψ+ ∼ ϕ1 and ψ− ∼ ϕ2 ). Physics of Atoms and Molecules 200 12 Molecules εH11 A2/ (H11-H22) H22 ε+ Abbildung 12.20. Energy splitting of the molecular orbital states of heteronuclear diatomic molecules. Abbildung 12.21. Correlation diagram between united atom and separated atom states for homonuclear diatomic molecules. The energy spacings are arbitrary and the actual energies vary from molecule to molecule. The diagonal lines sketch only trends and for a specic distance R the actual energies have to be calculated. Physics of Atoms and Molecules 12.6 Benzene: Application of the LCAO-MO method 201 12.6 Benzene: Application of the LCAO-MO method In molecules with more than two (H2+ ) identical atoms the symmetry is higher. In the case of benzene C6 H6 we nd six identical atoms with six single bonds (σ ) between the carbon atoms C and the hydrogen atoms H and six bonds between the carbon atoms. This is due to sp2 hybridization. One pz atomic orbital per carbon atom remains and is responsible for double bonds between the carbons. The molecule does not change under rotation about the z-axis (see Fig. 12.6) by 60◦ . Thus it has C6 symmetry. Due to the fact that all atoms are identical, the H 6 z y 5 1 4 2 x H H H H 3 H Abbildung 12.22. 'Resonance structure' of benzene. The 6 carbon atoms are equivalent and the molecule has C6 symmetry. The z-axis is directed perpendicular out of the paper plane. picture of a 'resonance structure' as presented in Fig. 12.6 has to be wrong. The molecular orbital wave functions are given by X ψ= ci φi (12.181) i with equal coecients |ci |. The probability to nd an electron at the atom i does not depend on i and the electron is called delocalized. There are six electrons from the six pz orbitals, which are delocalized. To determine the coecients we − use the symmetry property C6 of the system. Assume a pz atomic orbital φ(→ r) centered at the origin of the molecule. The atomic orbital centered at the atom i − → at R j is given by − → − → φj (→ r ) = φ(− r − R j ). (12.182) If H is the Hamilton operator describing the delocalized electrons, the operator commutes with R60 : [H, R60 ] = 0 (12.183) and both operators share the same eigenfunctions. The eigenvalues λ are → → → r ) = λψ(− r) R60 ψ(− r ) ≡ ψ(R60 − 6 λ = 1 2π λk = e i 6 k , k = 0, ±1, ±2. Physics of Atoms and Molecules (12.184) (12.185) (12.186) 202 12 Molecules If we let R60 operate on ψ we nd X X − → → − → → − cj φj (− r − R j) = cj R60 φj (→ r − R j) R60 ψ(− r ) = R60 = j X − → → cj φj−1 (− r − R j−1 ) j (12.187) j = λψ ⇒ = e k i 2π 6 cj φj−1 = e i 2π k 6 cm+1 φm = e i 2π k 6 X ⇒ X (12.188) − → → cj φj (− r − R j) (12.189) cj φj (12.190) j j j−1=m X X j X m φm linear independent (12.191) cm φm , m 2π cm+1 = ei 6 k cm cm+6 = cm . (12.192) (12.193) The solution is (k) cj (k) 2π (k) = ei 6 kj c0 (k) |cj | = |c0 |. (12.194) (12.195) The probability to nd an electron in the status k at the j th carbon atom is (k) proportional to ∝ |cj |2 . Since this probability is the same for every j the wave function has to be symmetric in j and is given by ψk = 6 X (k) → cj φj (− r 6 X 2π − → − → (k) → − R j ) = c0 ei 6 kj φj (− r − R j ). j=1 (12.196) j=1 (k) The √ coecient c0 in equation 12.196 is the normalization coecient (for example 1/ 6) and we can now solve the system of equations (k) (Hij − ²k Sij ) cj = 0 (12.197) and calculate the eigenenergies. We can set Hi,i = α (12.198) Hi,i±1 = β (12.199) which is independent of i and set which is the resonance integral and does not depend on i and is cyclic i−1 = i+5. In the Hückel method the following approximations are made: Physics of Atoms and Molecules 12.6 Benzene: Application of the LCAO-MO method 203 Only the nearest neighbor interactions are not zero Hij 6= 0, j =i±1 Overlap with the neighboring atoms are neglected Sij = δij Inserting the coecients into equations 12.197 we nd with the Hückel method (k) (k) (k) 0 = ( H11 −²k S11 )c1 + ( H12 −²k S12 )c2 + |{z} . . . +( H16 −²k S16 )c6 |{z} |{z} |{z} α =0 β β (12.200) i 2π k 6 i 2π k2 6 0 = (α − ²k )e i 2π k6 6 + βe + βe ³ 2π ´ 2π 0 = (α − ²k ) + β ei 6 k + e−i 6 k µ ¶ 2πk ²k = α + 2β cos , k = 0, ±1, ±2. 6 (12.201) (12.202) (12.203) The terms α = hi|H|ii and β = hi|H|i ± 1i are negative and β is the interaction between the nearest neighbors which result in a stabilizing interaction. The lowest energy term is ²0 = α + 2β , which is the ground state of benzene. According to Pauli's principle we can ll up the ground state with two electrons of opposite spin and go on with the second next lowest state and so on. Since we can ll up the states with 6 pz electrons the total ground state energy is (in the Hückel approximation without electron-electron interaction S) Eges ²0 ²1 ²2 ²3 = = = = = 2(α + 2β) + 4(α + β) = 6α + 8β α + 2β α+β α−β α − 2β. (12.204) (12.205) (12.206) (12.207) (12.208) The total energy of the six delocalized electrons is Eges = 6α + 8β , which is 2β lower than the total energy for three localized double bonds with E = 6α+6β (within the Hückel approximations). Therefore, the stabilizing of the delocalization of the electrons is 2β . The wave functions ψk are given without normalization ψ0 = φ1 + φ2 + φ3 + φ4 + φ5 + φ6 (12.209) ±iπ/3 ±i2π/3 ±iπ ±i4π/3 ±i5π/3 ±i6π/3 ψ±1 = e φ1 + e φ2 + e φ3 + e φ4 + e φ5 + e φ6 (12.210) ±i2π/3 ±i4π/3 ±i2π ±i8π/3 ±i10π/3 ±i12π/3 ψ±2 = e φ1 + e φ2 + e φ3 + e φ4 + e φ5 + e φ6 (12.211) ψ3 = −φ1 + φ2 − φ3 + φ4 − φ5 + φ6 (12.212) Physics of Atoms and Molecules 204 12 Molecules k=3 α − 2β k = -2 k=2 α-β α k = -1 k=1 α+β k=0 α + 2β Abbildung 12.23. Energy eigenvalues of the benzene molecule calculated with Hückel method. The wave function ψ0 has zero knots, ψ±1 has one knot line, ψ±2 has two knot lines, and ψ3 has three knot lines going through the molecule. 1 2 3 4 5 6 ψ0 ψ−1 ψ−2 ψ3 Abbildung 12.24. Qualitative representation of the benzene wave functions with increasing number of knots along the benzene ring. Physics of Atoms and Molecules 12.6 Benzene: Application of the LCAO-MO method 205 Abbildung 12.25. Molecular orbitals of benzene forming single bonds (a) and double bonds (b). Physics of Atoms and Molecules 206 12 Molecules 12.7 Hybridization, structure of polyatomic molecules Atomic orbital wave functions represent solutions of the SE. A linear superposition of solutions is again a solution of the SE. Molecular orbitals can be built of superpositions of atomic orbitals resulting in binding and antibinding orbitals (Fig. 12.7). For example upon excitation of an electron of the s-orbital of a carbon (a) Molecular orbitals combinations of s-orbitals (b) Molecular orbitals combinations of pz-orbitals (c) Molecular orbitals combinations of px and py-orbitals Abbildung 12.26. atom ((1s)2 (2s)2 (2p)2 ) to the excited state ((1s)2 (2s)(2p)3 ), we get one electron in the 2s, 2px , 2py , and 2pz orbital, and we can form a linear superposition of several wave functions (Fig. 12.7). This is called hybridization. Hybridization of an s-orbital and an pz -orbital leads to digonal hybridization with MO wave functions ψ+ and ψ+ (Fig. 12.7). These two functions give rise to two σ bonds in opposite directions along the z-axis. Ethine is an example for sp-hybridization (Fig. 12.7). 1 (ψs + ψpz ) N+ 1 = (ψs − ψpz ) N− ψ+ = (12.213) ψ− (12.214) (12.215) Physics of Atoms and Molecules 12.7 Hybridization, structure of polyatomic molecules 207 Hybridization of an s-orbital and two p-orbitals is called sp2 hybridization. This sp hybridization of s and pz orbitals Abbildung 12.27. Molecular orbitals of ethine: The sp-orbitals form the σ-bonds, while the remaining px and py orbitals form the π-bonds. Abbildung 12.28. type of hybridization is responsible for the molecular structure of ethylene (Fig. 12.7, Fig. 12.7). The angle between the three σ -bonds, which lie within one plane is 120 ◦ . ´ √ 1 ³ ψ1 = √ ψs + 2ψpx (12.216) 3 Ã ! r 1 1 3 ψp − √ ψpx ψ2 = √ ψs + (12.217) 2 y 3 2 Ã ! r 3 1 1 ψs − ψp − √ ψpx (12.218) ψ3 = √ 2 y 3 2 (12.219) Hybridization of an s-orbital and three p-orbitals is called sp3 hybridization. This type of hybridization is responsible for the molecular structure of methane (Fig. 12.7). The four new wave functions called hybrid orbitals are equivalent and Physics of Atoms and Molecules 208 12 Molecules (a) sp2 hybridized wave functions (b) pz wave functions Abbildung 12.29. wave function of the σ-electrons wave function of the π-electrons Abbildung 12.30. orthonormal (in the sense of the scalar product) and point in the four dierent directions of a tetrahedra: (1,1,1), (1,-1,-1), (-1,1,-1), and (-1,-1,1). The four unpaired electrons occupy the wave functions ψ1 ,..., ψ4 with parallel spins (Hund's rule) and experience a reduced Coulomb repulsion because of reduced overlap of the wave functions. The resulting angle between the main directions of the wave functions is 109.47 ◦ . For the directions we can write → ψ1 ∝ − r 1 =x+y+z − → ψ2 ∝ r 2 = −x − y + z − → → r 1·− r2 p cos θ = − → → 2 ( r ) (− r )2 1 = (12.220) (12.221) (12.222) 2 −1 − 1 + 1 1 √ √ = − = 109.47 ◦ . 3 3 3 Physics of Atoms and Molecules (12.223) 12.7 Hybridization, structure of polyatomic molecules 209 Abbildung 12.31. Atomic wave functions of the carbon atom (left) and its hybridized wave functions (right). The angle between the main directions of the wave functions is 109.47 ◦ . In the methane molecule (CH4 ) the four molecular orbitals responsible for the four single bonds between the carbon and the four hydrogen atoms are given by the hybrid wave functions ψi and the hydrogen wave functions ϕi (see Fig. 12.7) 1 + b2 1 = √ 2 a + b2 1 = √ a 2 + b2 1 = √ a 2 + b2 Ψ1 = √ Ψ2 Ψ3 Ψ4 a2 (aϕ1 + bψ1 ) (12.224) (aϕ2 + bψ2 ) (12.225) (aϕ3 + bψ3 ) (12.226) (aϕ4 + bψ4 ) . (12.227) Hybridization result in strongly increased electron overlap and therefore in stron- Abbildung 12.32. Hybridized wave functions and molecular orbitals of the methane molecule (left) and ethane molecule (right). ger chemical bonds. Physics of Atoms and Molecules 210 12 Molecules 12.8 Van der Waals interaction Absorption (a.u.) Emission intensity (a.u.) 12.9 Electronic transitions in molecules 200 250 300 W 350 400 450 500 550 avelength (nm) Abbildung 12.33. . Abbildung 12.34. Hybridized wave functions with sp2 hybridization and angle of 120 ween the orbitals (left). The remaining pz orbital (right). Physics of Atoms and Molecules ◦ bet- 12.9 Electronic transitions in molecules 211 Abbildung 12.35. Hybridized wave functions (sp2 ) and molecular orbitals of the ethylene molecule (upper panel). The remaining two pz orbital perpendicular to the σ bonds form a double bond (lower panel). Abbildung 12.36. Hybridized wave functions (sp). The remaining two py and two pz orbitals lead to two π bonds perpendicular to the σ bond. The molecules can form triple bonds (e.g. ethine). Physics of Atoms and Molecules 212 12 Molecules Abbildung 12.37. . Physics of Atoms and Molecules

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